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Irina18 [472]
3 years ago
5

The probability of finding electrons in certain regions of an atom is described by?

Chemistry
1 answer:
inysia [295]3 years ago
5 0

Answer:

orbits

Explanation:

An orbital is a region of space where there is a high probability of finding an electron.

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Consider the perbromate anion. What is the central atom? Enter its chemical symbol. How many lone pairs are around the central a
Yanka [14]

Answer:

See explanation

Explanation:

The central atom in the perbromate ion is bromine. The chemical symbol of bromine is Br. There are no lone pairs around the central bromine atom. The ion is tetrahedral in shape hence we expect a bond angle of 109°. 27 which is the ideal tetrahedral bond angle. The actual bond angle of the prebromate ion is 109.5°. The perbromate ion is BrO4^-

The observed bond angle is very close to the ideal value because of the absence of lone pairs of electrons from the central atom in the ion.

8 0
3 years ago
Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl. In an experiment 2
erastovalidia [21]

Answer:

Chloroform= limiting reactant

0.209mol of CCl4 is formed

And 32.186g of CCl4 is formed

Explanation:

The equation of reaction

CHCl3 + Cl2= CCl4 + HCl

From the equation 1 mol of

CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4

From the question

25g of CHCl3 really with Cl2

Molar mass of CHCl3= 119.5

Molar mass of Cl2 = 71

Hence moles of CHCl3= 25/119.5 = 0.209mol

Moles of Cl2 = 25/71 = 0.352mol

Hence CHCl3 is the limiting reactant

Since 1 mole of CHCl3 gave 1mol of CCl4

It implies that 0.209moles of CHCl3 will also give 0.209mol of CCl4

Mass of CCl4 formed = moles× molar mass= 0.209×154= 32.186g

6 0
3 years ago
At a certain temperature the vapor pressure of pure acetic acid HCH3CO2 is measured to be 226.torr. Suppose a solution is prepar
Rudiy27

Answer:

The partial pressure of acetic acid is 73.5 torr

Explanation:

Step 1: Data given

Total pressure is 226 torr

mass of acetic acid = 126 grams

mass of methanol = 141 grams

 

Step 2: Calculate moles of acetic acid

moles acetic acid = mass acetic acid / molar mass acetic acid

moles acetic acid = 127 grams / 60.05 g/mol

moles acetic acid = 2.115 moles

Step 3: Calculate moles of methanol

moles methanol = 141 grams / 32.04 g/mol

moles methanol = 4.40 moles

Step 4: Calculate total moles

Total moles = moles of acetic acid + moles methanol

Total moles = 2.115 moles + 4.40 moles

Total moles = 6.515 moles

Step 5: Calculate mole fraction of acetic acid

2.115 moles / 6.515 moles = 0.325

Step 6: Calculate partial pressure of acetic acid

P(acetic acid) = 0.325 * 226

P(acetic acid) = 73.45 torr ≈73.5

 

We can control this by calculating the partial pressure of methanol

mole fraction of methanol = (6.515-2.115)/6.515 = 0.675

P(methanol) = 0.675 * 226 = 152.55

226 - 152.55 = 73.45 torr  

The partial pressure of acetic acid is 73.5 torr

4 0
3 years ago
How many grams of the excess reactant remain assuming the reaction goes to completion and that you start with 15.5g of na2s and
tangare [24]
The reaction between Na2S and CuSO4 will give us the balanced chemical reaction of,
                                     Na2S + CUSO4 --> Na2SO4 + CuS
This means that for every 78g of Na2S, there needs to be 159.6 g of CuSO4. The ratio is equal to 0.4887 of Na2S: 1 of CuSO4. Thus, for every 12.1g of CuSO4, we need only 5.91 g of Na2S. Thus, there is an excess of 9.58 g of Na2S. The answer is letter C. 
8 0
3 years ago
Read 2 more answers
Given the standard heats of reaction
ANTONII [103]

Answer:

Explanation:

M(s) → M (g ) + 20.1 kJ --- ( 1 )

X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )

M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )

( 3 ) - 2 x ( 2 ) - ( 1 )

M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s)  → M X₄ (g ) - 98.7 kJ -  2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ

0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ

4 X (g ) +  M (g ) =  M X₄ (g ) - 773.4kJ

heat of formation of M X₄ (g ) is - 773.4 kJ

Bond energy of one M - X bond =  773.4 / 4 =  193.4 kJ / mole

6 0
3 years ago
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